# Evaluate $\int_{-\frac {\pi}{4}}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$

I am trying to evaluate the definite integral of (ex. 36 pg. 523, James Stewart Calculus 7e): $$\int_{-\frac {\pi}{4}}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$$ However, I have not got any possible approach recently. I have tried using trigonometric substitution but $$x^2$$ keep challenging me. I would be grateful if there is any suggestions, thank you!

• The integrand is an odd function. Sep 6 at 9:09

Take the function, I will call it $$f$$, and plug in $$-x$$ for $$x$$. What do you get? You get exactly $$-f(x)$$. Now notice that you are integrating over the interval $$(-\frac{\pi}{4},\frac{\pi}{4})$$. That which you will add in the positive region, you will substract by integrating in the negative region. Hence the result is $$0$$.

$$I=\displaystyle\int_{-\frac {\pi}{4}}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$$

$$I=\displaystyle\int_{-\frac {\pi}{4}}^{0} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx=-\int_{0}^{-\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$$

Let $$u=-x\implies du=-dx$$

$$\displaystyle I=-\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan (-u)}{1+{\cos^4{-u}}}(-du)+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$$

$$\displaystyle I=-\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan (u)}{1+{\cos^4{u}}}(du)+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$$

$$\displaystyle I=-\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan (x)}{1+{\cos^4{x}}}(dx)+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$$

$$I=0$$

The integrad is an odd function: $$f(-x)=-f(x)$$, so $$\int_{-\pi/4}^{\pi/4}f(x) dx=0.$$

• Did you check the comment? Sep 6 at 10:10

The function $$f(x)=\frac{x^2 \tan x}{1+\cos^4 x}$$ is odd. Hence any integral of the form $$\int_{-a}^{a} {f(x)} {dx}$$ is equal to zero.